Bài 2:

$a,n_{CuSO_4}=\dfrac{12}{160}=0,075(mol)\\ \Rightarrow n_{Cu}=n_S=0,075(mol);n_O=4n_{CuSO_4}=0,3(mol)\\ \Rightarrow m_{Cu}=0,075.64=4,8(g);m_S=0,075.32=2,4(g);m_O=0,3.16=4,8(g)\\ b,n_{NH_4NO_3}=\dfrac{20}{80}=0,25(mol)\\ \Rightarrow n_N=0,5(mol);n_H=1(mol);n_O=0,75(mol)\\ \Rightarrow m_N=0,5.14=7(g);m_H=1.1=1(g);m_O=0,75.16=12(g)\\ c,n_{H_2O}=\dfrac{9}{18}=0,5(mol)\\ \Rightarrow n_H=1(mol);n_O=0,5(mol)\\ \Rightarrow m_H=1.1=1(g);m_O=0,5.16=8(g)\\ d,n_{NaOH}=\dfrac{3,6.10^{23}}{6.10^{23}}=0,6(mol)\\ \Rightarrow n_{Na}=n_O=n_H=0,6(mol)\\ \Rightarrow m_{Na}=0,6.23=13,8;m_O=0,6.16=9,6(g);m_H=0,6.1=0,6(g)$

 

`#Ly`

$a)$

`n_(CuSO_4` `=` `(12)/(160)` `=` $0,075$ $(mol)$

Ta có:

`n_(Cu` `=` `n_S` `=` `n_(CuSO_4` `=` $0,075$ $(mol)$

Và `n_(O` `=` `4n_(CuSO_4` `=` $0,3$ $(mol)$

`m_(Cu` `=` $0,075.64=4,8$ $(g)$

`m_S` `=` $0,075.32=2,4$ $(g)$

`m_O` `=` $0,3.16=4,8$ $(g)$

$b)$

`n_(NH_4NO_3` `=` `(20)/(80)` `=` $0,25$ $(mol)$

Ta có:

`n_N` `=` `2n_(NH_4NO_3` `=` $0,5$ $(mol)$

`n_H` `=` `4n_(NH_4NO_3` `=` $1$ $(mol)$

`m_N` `=` `0,5.14=7` $(g)$

`m_H` `=` `1.1` `=` `1` $(g)$

`m_O` `=` `20-7-1=12` $(g)$

$c)$

`n_(H_2O` `=` `9/18` `=` $0,5$ $(mol)$

Ta có:

`n_O` `=` `n_(H_2O` `=` $0,5$ $(mol)$

`m_O` `=` $0,5.16=8$ $(g)$

`m_H` `=` `9-8=1` $(g)$

$d)$

`n_(NaOH` `=` `(3,6.10^23)/(6.10^23)` `=` `0,6` $(mol)$

Ta có:

`n_(Na` `=` `n_O` `=` `n_H` `=` `n_(NaOH` `=` `0,6` $(mol)$

`m_(Na` `=` `0,6.23=13,8` $(g)$

`m_(O` `=` `0,6.16=9,6` $(g)$

`m_H` `=` `0,6.1=0,6` $(g)$